Numerical Analysis

Numerical Analysis

Equations cannot always be solved algebraically, so it becomes necessary to use numerical analysis to find a very close approximation to the roots of that equation. There are several methods which can be used to find the solution which I will investigate. These methods are the change of sign Newton Raphson and the rearrangement method.

Change of sign

The decimal search method consists of testing different values for Ð''x' and seeing whether the sign is positive of negative.

There are a number of ways of applying the change of sign method to find the roots of an equation. I will apply the decimal search method.

This is done systematically, so you would test all of the figures to a certain degree of accuracy between too figures. So if you can see on a graph that the root of an equations is between 1 and 2, then you would test all the numbers between to 1 decimal place. Then you would find the two numbers it changes between and test all the numbers between those to 2 decimal places. The pattern is continued until an accurate root is found.

The equation I will use for this method is:

The graph of this equation is shown below.

The notation shows the interval between which the roots lie.

We know that the root lies between 1 and 2, so we test the values from 1.0 to 2.0.

x value y value

1.0 -5.5

1.1 -5.219

1.2 -4.872

1.3 -4.453

1.4 -3.956

1.5 -3.375

1.6 -2.704

1.7 -1.937

1.8 -1.068

1.9 -0.091

2.0 1

We can now see that the root lies between so a greater degree of accuracy is needed.

X value Y value

1.9 -0.091

1.91 0.012871

The sign changes between 1.90 and 1.91 so now we do it to 3 decimal places.

X value Y value

1.9 -0.091

1.901 -0.080664299

1.902 -0.070317192

1.903 -0.059958673

1.904 -0.049588736

1.905 -0.039207375

1.906 -0.028814584

1.907 -0.018410357

1.908 -0.007994688

1.909 0.002432429

So we can conclude that the root is 1.9085

Decimal search fails

The decimal search does not work in every situation. Sometimes it finds roots where there are none, or it fails to find a root.

For example if a graph changes sign twice in between the interval we test, no roots are found.

If we search for a root between there is no change of sign, and so the method fails.

It also fails when the graph contains asymptotes that do not cross the x axis.

The graph shows that there are no roots between however the method finds a root because the sign is positive at x=1 and negative at x=2.

Newton Raphson/ Fixed point iteration

Fixed point iteration involves repeating the same equation to increase the level of accuracy of finding the root of an equation. For the Newton Raphson method we use the Formula:

This means that the next iteration is given by the first iteration minus the equation divided by the differentiation of the equation.

When applying the Newton raphson method we start with an estimated root. We then draw a tangent from the curve at this point to find a more accurate root. We repeat this process until we achieve a suitable degree of accuracy.

The formula I will use is

The graph for this equation looks like this,

The estimated root I will use is 2. I will put this into my equation

In order to do this on my calculator, I must type in the following equation.

First I must press 2 and then the Ð''=' button. Then I type

ANS-((ANS4+3ANS2-11)/(4ANS3+6ANS))

This gives me the next iteration 1.614

This graph shows where the tangent has been drawn and how this is closer to the root.

I must press Ð''=' again to find the next iteration

This is 1.4781

I can repeatedly press the Ð''equals' button until I achieve a degree of accuracy that is suitable. I will use 5 decimal places.

Iteration X axis value

X1 1.61364

X2 1.47805

X3 1.46306

X4 1.46289

X5 1.46289

...